A - Bi-shoe and Phi-shoe(素数筛)

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在一个名为Xzhiland的地方，竹竿跳远是一项非常受欢迎的运动。为了满足学生的幸运数字要求，教练需要购买特定得分的竹子。通过计算每个学生幸运数字对应的最小素数长度的竹子总成本，来解决这个问题。

Bamboo Pole-vault is a massively popular sport in Xzhiland. And Master Phi-shoe is a very popular coach for his success. He needs some bamboos for his students, so he asked his assistant Bi-Shoe to go to the market and buy them. Plenty of Bamboos of all possible integer lengths (yes!) are available in the market. According to Xzhila tradition,

Score of a bamboo = Φ (bamboo's length)

(Xzhilans are really fond of number theory). For your information, Φ (n) = numbers less than n which are relatively prime (having no common divisor other than 1) to n. So, score of a bamboo of length 9 is 6 as 1, 2, 4, 5, 7, 8 are relatively prime to 9.

The assistant Bi-shoe has to buy one bamboo for each student. As a twist, each pole-vault student of Phi-shoe has a lucky number. Bi-shoe wants to buy bamboos such that each of them gets a bamboo with a score greater than or equal to his/her lucky number. Bi-shoe wants to minimize the total amount of money spent for buying the bamboos. One unit of bamboo costs 1 Xukha. Help him.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 10000) denoting the number of students of Phi-shoe. The next line contains n space separated integers denoting the lucky numbers for the students. Each lucky number will lie in the range [1, 106].

Output

For each case, print the case number and the minimum possible money spent for buying the bamboos. See the samples for details.

Sample Input

3

5

1 2 3 4 5

6

10 11 12 13 14 15

2

1 1

Sample Output

Case 1: 22 Xukha

Case 2: 88 Xukha

找到大于其本身的最小的素数，求和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
bool p[1234567];
void isprime()//素数筛
{
   memset(p, true, sizeof(p));//初始化
   p[1] = false;//1不是素数
   for(int i=2;i<1234567;i++)
   {
      if(p[i])
      {
         for(int j=2*i;j<1234567;j+=i)//对于其倍数进行同样的操作
         {
            p[j] = false;
         }
      }
   }
}
int main()
{
  int T;
  isprime();
  scanf("%d", &T);
  for(int t=1;t<=T;t++)
  {
     int n, m;
     scanf("%d", &n);
      long long int k = 0;//求和，注意使用long long 
     for(int i=1;i<=n;i++)
     {
       scanf("%d", &m);
       for(int j=m+1;j<1234567;j++)
       {
         if(p[j])
         {
           k += j;
           break;
         }
       }
     }
     printf("Case %d: %lld Xukha\n", t, k);
  }
  return 0;
}